1 Zinc reacts with acids to produce H 2 gas.Zinc reacts with acids to produce H 2 gas. Have 10.0 g...
-
Upload
clifton-thomas -
Category
Documents
-
view
226 -
download
2
Transcript of 1 Zinc reacts with acids to produce H 2 gas.Zinc reacts with acids to produce H 2 gas. Have 10.0 g...
1
• Zinc reacts with Zinc reacts with acids to produce acids to produce HH22 gas. gas.
• Have 10.0 g of ZnHave 10.0 g of Zn
• What volume of What volume of 2.50 M HCl is 2.50 M HCl is needed to convert needed to convert the Zn completely?the Zn completely?
SOLUTION SOLUTION STOICHIOMETRYSTOICHIOMETRY
Section 5.10Section 5.10
2GENERAL PLAN FOR GENERAL PLAN FOR STOICHIOMETRY CALCULATIONSSTOICHIOMETRY CALCULATIONS
GENERAL PLAN FOR GENERAL PLAN FOR STOICHIOMETRY CALCULATIONSSTOICHIOMETRY CALCULATIONS
Mass zinc
StoichiometricfactorMoles
zincMoles HCl
Mass HCl
VolumeHCl
3
Step 1: Step 1: Write the balanced equationWrite the balanced equation
Zn(s) + 2 HCl(aq) --> ZnClZn(s) + 2 HCl(aq) --> ZnCl22(aq) + H(aq) + H22(g)(g)
Step 2: Step 2: Calculate amount of ZnCalculate amount of Zn
10.0 g Zn • 1.00 mol Zn65.39 g Zn
= 0.153 mol Zn10.0 g Zn • 1.00 mol Zn65.39 g Zn
= 0.153 mol Zn
Zinc reacts with acids to produce HZinc reacts with acids to produce H22 gas. If you gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely?needed to convert the Zn completely?
Zinc reacts with acids to produce HZinc reacts with acids to produce H22 gas. If you gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely?needed to convert the Zn completely?
Step 3: Step 3: Use the stoichiometric factorUse the stoichiometric factor
4
Step 3: Step 3: Use the stoichiometric factorUse the stoichiometric factor
0.153 mol Zn • 2 mol HCl1 mol Zn
= 0.306 mol HCl0.153 mol Zn • 2 mol HCl1 mol Zn
= 0.306 mol HCl
0.306 mol HCl • 1.00 L
2.50 mol = 0.122 L HCl0.306 mol HCl •
1.00 L2.50 mol
= 0.122 L HCl
Zinc reacts with acids to produce HZinc reacts with acids to produce H22 gas. If you gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely?needed to convert the Zn completely?
Zinc reacts with acids to produce HZinc reacts with acids to produce H22 gas. If you gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely?needed to convert the Zn completely?
Step 4: Step 4: Calculate volume of HCl req’dCalculate volume of HCl req’d
5
ACID-BASE REACTIONSACID-BASE REACTIONSTitrationsTitrations
ACID-BASE REACTIONSACID-BASE REACTIONSTitrationsTitrations
HH22CC22OO44(aq) + 2 NaOH(aq) --->(aq) + 2 NaOH(aq) --->
acidacid basebase
NaNa22CC22OO44(aq) + 2 H(aq) + 2 H22O(liq)O(liq)
Carry out this reaction using a Carry out this reaction using a TITRATIONTITRATION..
Oxalic acid,Oxalic acid,
HH22CC22OO44
6Setup for titrating an acid with a baseSetup for titrating an acid with a base
CCR, page 186
7
TitratioTitrationn
TitratioTitrationn
1. Add solution from the 1. Add solution from the buret.buret.
2. Reagent (base) reacts 2. Reagent (base) reacts with compound (acid) in with compound (acid) in solution in the flask.solution in the flask.
3. Indicator shows when 3. Indicator shows when exact stoichiometric exact stoichiometric reaction has occurred.reaction has occurred.
4. Net ionic equation4. Net ionic equation
HH++ + OH + OH-- --> H --> H22OO
5. At equivalence point 5. At equivalence point
moles Hmoles H++ = moles OH = moles OH--
8
1.065 g of H1.065 g of H22CC22OO44 (oxalic (oxalic
acid) requires 35.62 mL acid) requires 35.62 mL
of NaOH for titration to of NaOH for titration to
an equivalence point. an equivalence point.
What is the concentra-What is the concentra-
tion of the NaOH?tion of the NaOH?
LAB PROBLEM #1: Standardize a LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately solution of NaOH — i.e., accurately determine its concentration.determine its concentration.
LAB PROBLEM #1: Standardize a LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately solution of NaOH — i.e., accurately determine its concentration.determine its concentration.
9
1.065 g of H1.065 g of H22CC22OO44 (oxalic acid) requires 35.62 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH?point. What is the concentration of the NaOH?
1.065 g of H1.065 g of H22CC22OO44 (oxalic acid) requires 35.62 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH?point. What is the concentration of the NaOH?
Step 1: Step 1: Calculate amount of HCalculate amount of H22CC22OO44
1.065 g • 1 mol
90.04 g = 0.0118 mol1.065 g •
1 mol90.04 g
= 0.0118 mol
0.0118 mol acid • 2 mol NaOH1 mol acid
= 0.0236 mol NaOH
Step 2: Step 2: Calculate amount of NaOH req’dCalculate amount of NaOH req’d
10
1.065 g of H1.065 g of H22CC22OO44 (oxalic acid) requires 35.62 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH?point. What is the concentration of the NaOH?
1.065 g of H1.065 g of H22CC22OO44 (oxalic acid) requires 35.62 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH?point. What is the concentration of the NaOH?
Step 1: Step 1: Calculate amountCalculate amount of Hof H22CC22OO44
= 0.0118 mol acid= 0.0118 mol acid
Step 2: Step 2: Calculate amount of NaOH req’dCalculate amount of NaOH req’d
= 0.0236 mol NaOH= 0.0236 mol NaOH
Step 3: Step 3: Calculate concentration of NaOHCalculate concentration of NaOH
0.0236 mol NaOH0.03562 L
0.663 M0.0236 mol NaOH
0.03562 L 0.663 M
[NaOH] = 0.663 M[NaOH] = 0.663 M
11
LAB PROBLEM #2: LAB PROBLEM #2: Use standardized NaOH to determine Use standardized NaOH to determine the amount of an acid in an unknown.the amount of an acid in an unknown.
LAB PROBLEM #2: LAB PROBLEM #2: Use standardized NaOH to determine Use standardized NaOH to determine the amount of an acid in an unknown.the amount of an acid in an unknown.
Apples contain malic acid, CApples contain malic acid, C44HH66OO55..
CC44HH66OO55(aq) + 2 NaOH(aq) --->(aq) + 2 NaOH(aq) --->
NaNa22CC44HH44OO55(aq) + 2 H(aq) + 2 H22O(liq)O(liq)
76.80 g of apple requires 34.56 mL of 0.663 76.80 g of apple requires 34.56 mL of 0.663
M NaOH for titration. What is weight % of M NaOH for titration. What is weight % of
malic acid?malic acid?
12
76.80 g of apple requires 34.56 mL of 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is 0.663 M NaOH for titration. What is weight % of malic acid?weight % of malic acid?
76.80 g of apple requires 34.56 mL of 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is 0.663 M NaOH for titration. What is weight % of malic acid?weight % of malic acid?
Step 1: Step 1: Calculate amount of NaOH used.Calculate amount of NaOH used.
C • V = (0.663 M)(0.03456 L) C • V = (0.663 M)(0.03456 L)
= 0.0229 mol NaOH= 0.0229 mol NaOH
Step 2: Step 2: Calculate amount of acid titrated.Calculate amount of acid titrated.
0.0229 mol NaOH • 1 mol acid
2 mol NaOH
= 0.0115 mol acid= 0.0115 mol acid
13
76.80 g of apple requires 34.56 mL of 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is 0.663 M NaOH for titration. What is weight % of malic acid?weight % of malic acid?
76.80 g of apple requires 34.56 mL of 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is 0.663 M NaOH for titration. What is weight % of malic acid?weight % of malic acid?
Step 3: Step 3: Calculate mass of acid titrated.Calculate mass of acid titrated.
0.0115 mol acid • 134 gmol
= 1.54 g
Step 1: Step 1: Calculate amount of NaOH used. Calculate amount of NaOH used.
= 0.0229 mol NaOH= 0.0229 mol NaOH
Step 2: Step 2: Calculate amount of acid titratedCalculate amount of acid titrated
= 0.0115 mol acid= 0.0115 mol acid
14
76.80 g of apple requires 34.56 mL of 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is 0.663 M NaOH for titration. What is weight % of malic acid?weight % of malic acid?
76.80 g of apple requires 34.56 mL of 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is 0.663 M NaOH for titration. What is weight % of malic acid?weight % of malic acid?
Step 1: Step 1: Calculate amount of NaOH used. Calculate amount of NaOH used.
= 0.0229 mol NaOH= 0.0229 mol NaOH
Step 2: Step 2: Calculate amount of acid titratedCalculate amount of acid titrated
= 0.0115 mol acid= 0.0115 mol acid
Step 3: Step 3: Calculate mass of acid titrated.Calculate mass of acid titrated.
= 1.54 g acid= 1.54 g acid
Step 4: Step 4: Calculate % malic acid.Calculate % malic acid.1.54 g76.80 g
• 100% = 2.01%